1. Use the Graphical Method to solve the following LP Problem.
Maximize Z = 15x1 + 10x2
Subject to constraints,
(i) 4x1 + 6x2 ≤ 360
(ii) 3x1 + 0x2 ≤ 180
(iii) 0x1 + 5x2 ≤ 200
And
x1, x2 ≥ 0
Solution:
The given LP problem is already in mathematical form
Converting the inequality equation to equality, the constraints will be re-written as
4x1 + 6x2 = 360------------(1)
3x1 + 0x2 = 180------------(2)
0x1 + 5x2 = 200------------(3)
Solving equation (1)
4x1 + 6x2 = 360
Put x2 = 0 Put x1 = 0
4x1 = 360 6x2 = 360
X1 = 90 x2 = 60
So the points are (90,0) & (60,0)
Solving equation (2)
3x1 + 0x2 = 180 3x1 = 180
3x1 = 180
x1 = 60
Solving equation (3)
0x1 + 5x2 = 200
5x2 = 200
x2 = 40
Draw the Graph with the following points or equation
To Find the Feasible Region
Examine whether the origin (0, 0) satisfies the constraint. If it does, then all points on and below the
constraint equation towards the origin are feasibility region
4x1 + 6x2 ≤ 360
4*0+6*0 = 0 ≤ 360
If the origin (0, 0) does not satisfies the constraint, then all points on and above the constraint
equation away from the origin are feasibility points
Extreme Points Co-ordinates (x1, x2) Objective Function
Max Z = 15x1 + 10x2
O (0, 0) 15*0 + 10*0 = 0
A (60, 0) 15*60 + 10*0 = 900
B (60, 20) 15*60 + 10*20 = 1,100
C (30, 40) 15*30 + 10*40 = 850
D (0, 40) 15*0 + 10*40 = 400
Find out the maximum value from the above table, the maximum value of Z = 1,100 is achieved at the
point extreme B (60, 20)
Hence the optimal solution to the given LP problem is
x1 = 60 : x2 = 20
Max Z = 1,100
2. Use the Graphical Method to solve the following LP Problem.
Maximize Z = 60x1 + 40x2
Subject to constraints:
(i) x1 ≤ 25
(ii) x2 ≤ 35
(iii) 2x1 + x2 ≤ 60
And
x1, x2 ≥ 0
Solution
The given LP problem is already in mathematical form
Converting the inequality equation to equality, the constraints will be re-written as
x1 = 25--------------------(1)
x2 = 35--------------------(2)
2x1 + x2 = 60-------------(3)
Solving equation (1)
x1 = 25
Solving equation (2)
x2 = 35
Solving equation (3)
2x1 + x2 =60
Put x2 = 0 Put x1 = 0
2x1 = 60 x2 = 60
x1 = 30
So the points are (30,0) & (0,60)
Draw the Graph with the following points or equation
To Find the Feasible Region
Examine whether the origin (0, 0) satisfies the constraint. If it does, then all points on and below the
constraint equation towards the origin are feasibility points
If the origin (0, 0) does not satisfies the constraint. If it does, then all points on and above the
constraint equation away from the origin are feasibility points
Extreme Points Co-ordinates (x1, x2) Objective Function
Max Z = 60x1 + 40x2
O (0, 0) 60*0 + 40*0 = 0
A (0, 35) 60 *0 + 40*35 = 1,400
B (12.5, 35) 60 *12.5 + 40*35 = 2,150
C (25, 10) 60 *25 + 40*10 = 1,900
D (25, 0) 60*25 + 40*0 = 1,500
Find out the maximum value from the above table, the maximum value of Z = 2,150 is achieved at the
point extreme B (12.5, 35)
Max Z = 2,150
x1 = 12.5
x2 = 35
3. Use the Graphical Method to solve the following LP Problem.
Minimize Z = 3x1 + 2x2
Subject to the Constraints,
(i) 5x1 + x2 ≥ 10
(ii) x1 + x2 ≥ 6
(iii) x1 + 4x2 ≥ 12
And
x1, x2 ≥ 0
Solution
The given LP problem is already in mathematical form
Converting the inequality equation to equality, the constraints will be re-written as
5x1 + x2 = 10----------------(1)
x1 + x2 = 6-------------------(2)
x1 + 4x2 = 12----------------(3)
Solving equation (1)
5x1 + x2 = 10
Put x2 = 0, Put x1 = 0
5x1 = 10 x2 = 10
x1 = 2
So the points are (2, 10)
Solving equation (2)
x1 + x2 = 6
Put x2 = 0 put x1 = 0
x1 = 6 x2 = 6
So the points are (6, 6)
Solving equation (3)
x1 + 4x2 = 12
put x2 = 0 put x1 = 0
x1 = 12 4x2 = 12
x2 = 3
So the points are (12, 3)
Draw the Graph with the following points or equation
To Find the Feasible Region
Examine whether the origin (0, 0) satisfies the constraint. If it does, then all points on and below the
constraint equation towards the origin are feasibility points
If the origin (0, 0) does not satisfies the constraint. If it does, then all points on and above the
constraint equation away from the origin are feasibility points
Extreme Points Co-ordinates (x1, x2) Objective Function
Min Z = 3x1 + 2x2
A (12, 0) 3*12 + 2*0 = 36
B (4, 2) 3*4 + 2*2 = 16
C (1, 5) 3*1 + 2*5 = 13
D (0, 10) 3*0 + 2*10 = 20
Find out the minimum value from the above table, the minimum value of Z = 13 is achieved at the point
extreme C (1, 5)
Min Z = 13
x1 = 1
x2 = 5
4. Use the Graphical Method to solve the following LP Problem.
Maximize Z = 2x1 + x2
Subject to the constraints
(i) x1 + 2x2 ≤ 10
(ii) x1 + x2 ≤ 6
(iii) x1 - x2 ≤ 2
(iv) x1 – 2x2 ≤ 1
And
x1, x2≥ 0
Solution
The given LP problem is already in mathematical form
Converting the inequality equation to equality, the constraints will be re-written as
x1 + 2x2 = 10------------------- (1)
x1 + x2 = 6---------------------- (2)
x1 - x2 = 2----------------------- (3)
x1 – 2x2 = 1--------------------- (4)
Solving equation (1)
x1 + 2x2 = 10 put x1 = 0
put x2 = 0 2x2 = 10
x1 = 10 x2 = 5
So the points are (10, 5)
Solving equation (2)
x1 + x2 = 6
put x2 = 0 put x1 = 0
x1 = 6 x2 = 6
So the points are (6, 6)
Solving equation (3)
x1 - x2 = 2
put x2 = 0 put x1 = 0
x1 = 2 -x2 = 2
x2 = -2
So the points are (2, -2)
Solving equation (4)
x1 – 2x2 = 1
put x2 = 0 put x1 = 0
x1 = 1 -2x2 = 1
x2 = -1/2 = -0.5
So the points are (1, -0.5)
Draw the Graph with the following points or equation
Extreme Points Co-ordinates (x1, x2) Objective Function
Max Z = 2x1 + x2
O (0, 0) 2*0 + 0 = 0
A (1, 0) 2*1 + 0 = 2
B (3, 1) 2*3 + 1 = 7
C (4, 2) 2*4 + 2 = 10
D (2, 4) 2*2 + 4 = 8
E (0, 5) 2*0 + 5 = 5
Find out the maximum value from the above table, the maximum value of Z = 10 is achieved at the point
extreme C (4, 2)
Max Z = 10
x1 = 4
x2 = 2
5. Use the Graphical Method to solve the following LP Problem.
Minimize Z = -x1 + 2x2
Subject to the Constraints
(i) –x1 + 3x2 ≤ 10
(ii) x1 + x2 ≤ 6
(iii) x1 – x2 ≤ 2
And
x1, x2 ≥ 0
Solution
The given LP problem is already in mathematical form
Converting the inequality equation to equality, the constraints will be re-written as
–x1 + 3x2 = 10-----------------(1)
x1 + x2 = 6----------------------(2)
x1 – x2 = 2----------------------(3)
Solving equation (1)
–x1 + 3x2 = 10
Put x2 = 0 put x1 = 0
-x1 = 10 3x2 = 10
x1 = -10 x2 = 10/3 = 3.33
So the points are (-10, 3.33)
Solving equation (2)
x1 + x2 = 6
put x2 = 0 put x1 = 0
x1 = 6 x2 = 6
So the points are (6, 6)
Solving equation (3)
x1 – x2 = 2
put x2 = 0 put x1 = 0
x1 = 2 -x2 = 2
x2 = -2
So the points are (2, -2)
Draw the Graph with the following points or equation
Extreme Points Co-ordinates (x1, x2) Objective Function
Min Z = -x1 + 2x2
O (0, 0) -0 + 2*0 = 0
A (2, 0) -2 + 2*0 = -2
B (4, 2) -4 + 2*2 = 0
C (2, 4) -2 + 2*4 = 6
D (0, 3.33) -0 + 2*3.33 = 6.66
Find out the minimum value from the above table, the minimum value of Z = -2 is achieved at the point
extreme A (2, 0)
Min Z = -2
x1 = 2
x2 = 0
